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I am using replace function to replace numbers with characters in a list but seem to need to add an extra value to the index list for the code to work. (See example)

Simple version of by problem is that I have a list (0,1) and I want to replace 0 with "a" and 1 with "b". 

replace(0:1, 0:1, c("a","b"))

Generates following results

[1] "a" "1"
Warning message:
In x[list] <- values :
  number of items to replace is not a multiple of replacement length

when extend index list to 0:2 I get the desired output

> replace(0:1, 0:2, c("a","b"))
[1] "a" "b"

I obviously don't understand how the replace function works but I can't seem to figure out what I am missing.

Toph Bei Fong
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    The second argument to `replace` is a list of *indices* (not values) and R uses 1-based indexing. `0:1` only contains 1 valid index. – John Coleman Jun 10 '19 at 01:41
  • @JohnColeman - in the answer box below ↓ ;-) – thelatemail Jun 10 '19 at 01:49
  • @thelatemail I didn't want to simply diagnose the problem but rather wanted to suggest a workaround for what OP was trying to do. Having said that, I think that I do sometimes fall into the habit of answering questions in the comments. Less typing. – John Coleman Jun 10 '19 at 11:25

1 Answers1

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The second argument to replace is a list of indices (not values) and R uses 1-based indexing. 0:1 only contains 1 valid index. Thus in effect, 

replace(0:1, 0:1, c("a","b"))

is equivalent to

replace(0:1, 1, c("a","b"))

which generates the warning message.

It seems that you are really looking for the functionality of mapvalues from the plyr package. For example, 

x <- c(2,3,5,7,11)
plyr::mapvalues(x,c(3,7),c(10,12))

which leads to the output:

[1]  2 10  5 12 11

mapvalues is an example of a plyr function that doesn't have a direct analog in the newer package dplyr. See this question. Using both packages simultaneously is not recommended. This is why I used the more explicit plyr::mapvalues() rather than first library(plyr) and then simply mapvalues().

John Coleman
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